Proof.Under (), the first conclusion follows immediately. The second conclusion holds by invoking Attouch's theorem; see [1]}.
It applies because \(f^\nu \)  being proper, lsc, and convex implies that \(f\)  is lsc and convex by virtue of being the limit of \(f^\nu \) . Moreover, \(\operatorname{epi}f \ne \emptyset \)  because \(\varphi \)  is proper. The possibility \(f(u,x) = -\infty \)  is ruled about because it would have implied that \(f\)  is nowhere finite. Thus, \(f\)  is also proper.
\(\hrule height1pt \hbox{\vrule width1pt height5pt\hspace{5.0pt}\vrule width1pt} \hrule height1pt\)\(\)
